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Anni [7]
3 years ago
15

Qu 4.

Chemistry
1 answer:
Anna11 [10]3 years ago
3 0

Answer:

It's the second one down.

Explanation:

Gold

Mass 197

Number of Protons: 79

Number of Neutrons: 197 - 79 = 118

Number of electrons: = number of protons = 97

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For 520.0 ml of a buffer solution that is 0.175 m in hc2h3o2 and 0.165 m in nac2h3o2, calculate the initial ph and the final ph
alisha [4.7K]
KWIJRJHO0907 X 023KO
3 0
2 years ago
Dissolved hydrofluoric acid reacts with dissolved sodium hydroxide to form water and aqueous sodium fluoride
GalinKa [24]

Answer:

HF(aq)+NaOH(aq)→NaF(aq)+H2O(l)

Explanation:

Complete question

Dissolved hydrofluoric acid reacts with dissolved sodium hydroxide to form water and aqueous sodium fluoride. What is the net ionic equation

Equilibrium equation between the undissociated acid and the dissociated ions

HF(aq)⇌H+(aq)+F−(aq)

Sodium hydroxide will dissociate aqueous solution to produce sodium cations, Na+, and hydroxide anions, OH−

NaOH(aq)→Na+(aq)+OH−(aq)

Hydroxide anions and the hydrogen cations will neutralize each other to produce water.

H+(aq)+OH−(aq)→H2O(l)

On combining both the equation, we get –  

HF(aq)+Na+(aq)+OH−(aq)→Na+(aq)+F−(aq)+H2O(l)

The Final equation is  

HF(aq)+NaOH(aq)→NaF(aq)+H2O(l)

5 0
2 years ago
Classify the following reaction.
Klio2033 [76]

Answer:

Explanation:

combustion  reacts with oxygen

7 0
3 years ago
One way of obtaining pure sodium carbonate is through the decomposition of the mineral trona, Na3(CO3)(HCO3)·2H2O. 2Na3(CO3)(HCO
zhenek [66]
Percentage yield = (actual yield / theoretical yield) x 100%

The balanced equation for the decomposition is,
 2Na₃(CO₃)(HCO₃)·2H₂O(s) → 3Na₂CO₃(s) + CO₂(g) + 5H₂<span>O(g)

The stoichiometric ratio between </span>Na₃(CO₃)(HCO₃)·2H₂O(s)  and Na₂CO₃(s) is 2 : 3

The decomposed mass of Na₃(CO₃)(HCO₃)·2H₂O(s) = 1000 kg
                                                                                     = 1000 x 10³ g

Molar mass of Na₃(CO₃)(HCO₃)·2H₂O(s) = 226 g mol⁻¹
moles of Na₃(CO₃)(HCO₃)·2H₂O(s) = mass / molar mass
                                                         = 1000 x 10³ g / 226 g mol⁻¹
                                                         = 4424.78 mol

Hence, moles of Na₂CO₃ formed = 4424.78 mol x \frac{3}{2}
                                                     = 6637.17 mol

Molar mass of Na₂CO₃ = 106 g mol⁻¹

Hence, mass of Na₂CO₃ = 6637.17 mol x 106 g mol⁻¹
                                        = 703540.02 g
                                        = 703.540 kg

Hence, the theoretical yield of Na₂CO₃ =  703.540 kg
Actual yield of Na₂CO₃ = 650 kg

Percentage yield = (650 kg / 703.540 kg) x 100%
                            = 92.34%
7 0
3 years ago
2. The density of helium is 1.78 X 104 g/cm. What is this<br> density in Dg/um??
Zepler [3.9K]

Answer:

d=1.78\times 10^{-7}\ Dg/\mu m^3

Explanation:

Given,

The density of Helium is 1.78\times 10^4\ g/cm^3

We need to find the density in Dg/μm

We know that,

1 g = 10 dg

1 cm³ = 10¹² μm³

So,

d=1.78 \times 10^4\ g/cm^3\\\\=1.78 \times 10^4\times \dfrac{10\ dg}{10^{12}\ \mu m^3}\\\\=1.78\times 10^{-7}\ Dg/\mu m^3

So, the density of Helium is equal to 1.78\times 10^{-7}\ Dg/\mu m^3.

4 0
2 years ago
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